I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
Let the triangle be $ABC$ and external angle bisector of $\angle ABC$ cut $AC$ in $X$, of $\angle ACB$ cut $AB$ in $Y$, of $\angle BAC$ cut $BC$ in $Z$.
By angle bisector theorem,
$$\frac{AX}{XC}=-\frac{AB}{BC}... (1)$$
$$\frac{CZ}{ZB}=-\frac{CA}{AB}... (2)$$
$$\frac{BY}{YA}=-\frac{BC}{CA}... (3)$$
(1) ×(2) ×(3) gives,
$$\frac{AX.CZ.BY}{XC.ZB.YA}=-1$$
Therefore by converse of Menelaus Theorem X, Y, Z are collinear.
Best Answer
First, you prove that external bisectors $AD$, $BD$, and inner bisector $CD$ intersect in on point in the same manner as with 3 inner bisectors. => Points $C$, $I$ and $D$ lie on the same line.
Quadrilateral $IADB$ has two opposite angles $\angle IAD=\angle IBD=90^\circ$, so $IABD$ is cyclic with the center of circle $O$ in the center of $ID$. => Points $O$, $I$ and $D$ lie on the same line.
Then you combine both statements together.